Given the function f, of, x, equals, x, cubed, minus, 6, x, squared, minus, 36, x, commaf(x)=x 
3
 −6x 
2
 −36x, determine all intervals on which ff is both increasing and concave down.

To determine where the function \( f(x) = x^3 - 6x^2 - 36x \) is both increasing and concave down, we need to:

1. **Find the intervals where the function is increasing**: A function is increasing where its first derivative, \( f'(x) \), is positive.
2. **Find the intervals where the function is concave down**: A function is concave down where its second derivative, \( f''(x) \), is negative.

### Step 1: Find \( f'(x) \) and determine where it is positive.
Given:
\[
f(x) = x^3 - 6x^2 - 36x
\]
The first derivative is:
\[
f'(x) = 3x^2 - 12x - 36
\]
Now, solve \( f'(x) > 0 \) to find the intervals where the function is increasing:
\[
3x^2 - 12x - 36 > 0
\]
Divide the inequality by 3:
\[
x^2 - 4x - 12 > 0
\]
Factor or use the quadratic formula to solve:
\[
x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-12)}}{2(1)}
\]
\[
x = \frac{4 \pm \sqrt{16 + 48}}{2} = \frac{4 \pm \sqrt{64}}{2} = \frac{4 \pm 8}{2}
\]
Thus, the solutions are:
\[
x = \frac{4 + 8}{2} = 6 \quad \text{and} \quad x = \frac{4 - 8}{2} = -2
\]
So, the critical points are \( x = -2 \) and \( x = 6 \). To determine the intervals where \( f'(x) > 0 \), we test the sign of \( f'(x) \) in the intervals \( (-\infty, -2) \), \( (-2, 6) \), and \( (6, \infty) \).

- For \( x \in (-\infty, -2) \), choose \( x = -3 \):
  \[
  f'(-3) = 3(-3)^2 - 12(-3) - 36 = 27 + 36 - 36 = 27 > 0
  \]
  So, \( f'(x) > 0 \) in \( (-\infty, -2) \).
  
- For \( x \in (-2, 6) \), choose \( x = 0 \):
  \[
  f'(0) = 3(0)^2 - 12(0) - 36 = -36 < 0
  \]
  So, \( f'(x) < 0 \) in \( (-2, 6) \).

- For \( x \in (6, \infty) \), choose \( x = 7 \):
  \[
  f'(7) = 3(7)^2 - 12(7) - 36 = 147 - 84 - 36 = 27 > 0
  \]
  So, \( f'(x) > 0 \) in \( (6, \infty) \).

Thus, \( f(x) \) is increasing on the intervals \( (-\infty, -2) \) and \( (6, \infty) \).

### Step 2: Find \( f''(x) \) and determine where it is negative.
The second derivative is:
\[
f''(x) = 6x - 12
\]
Solve \( f''(x) < 0 \):
\[
6x - 12 < 0
\]
\[
x < 2
\]
Thus, \( f(x) \) is concave down for \( x < 2 \).

### Step 3: Determine the intervals where \( f(x) \) is both increasing and concave down.
From Step 1, \( f(x) \) is increasing on \( (-\infty, -2) \) and \( (6, \infty) \). From Step 2, \( f(x) \) is concave down for \( x < 2 \). The overlap of these conditions occurs in the interval \( (-\infty, -2) \).

### Conclusion:
The function \( f(x) = x^3 - 6x^2 - 36x \) is both increasing and concave down on the interval \( (-\infty, -2) \).

Would you like a deeper explanation on any part of the process?

Here are 5 related questions to explore further:
1. How do you find intervals of concavity for more complex functions?
2. What is the significance of points where the second derivative equals zero?
3. Can you explain how the first and second derivatives together define the shape of a curve?
4. How can the derivative test help in optimization problems?
5. What is the role of inflection points in concavity analysis?

**Tip:** Always remember that concavity helps describe how a function "bends." When concave down, a function curves downward like a frown!

Given the function f(x) = x^3 - 6x^2 - 36x, determine all intervals on which f is both increasing and concave down.

Explore the detailed process of finding intervals where f(x) = x^3 - 6x^2 - 36x is both increasing and concave down. Learn how to apply the first and second derivative tests to analyze the behavior of this cubic function. Suitable for students in Grades 11-12 or early college.

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